equation of plane

Let the length of the position vector be rrr and the angles formed by the vector with the positivecoordinate axes αα\alpha, ββ\beta, γγ\gamma. Let P=(x,y,z)PxyzP=(x,\,y,\,z) be an arbitrary point. Then PPP is in the plane ττ\tauiff its projection on the lineO⁢QOQOQ coincides with QQQ, i.e. iff the projection of the coordinate way of PPP is rrr. This may be expressed as the equationx⁢cos⁡α+y⁢cos⁡β+z⁢cos⁡γ=rxαyβzγrx\cos\alpha+y\cos\beta+z\cos\gamma=r or

between the variablesxxx, yyy, zzzrepresents always a plane. In fact, we may without hurting generality suppose that D≦0D0D\leqq 0. Now R:=A2+B2+C2>0assignRsuperscriptA2superscriptB2superscriptC20R:=\sqrt{A^{2}+B^{2}+C^{2}}>0. Thus the length of the radius vector of the point (A,B,C)ABC(A,\,B,\,C) is RRR. Let the angles formed by the radius vector with the positive coordinate axes be αα\alpha, ββ\beta, γγ\gamma. Then we can write

where DR≦0DR0\frac{D}{R}\leqq 0. The last equation represents a plane whose distance from the origin is -DRDR-\frac{D}{R} and whose normal line forms the angles αα\alpha, ββ\beta, γγ\gamma with the coordinate axes.

The plane can be represented also in a vectoral form, by using the position vector r→0subscriptnormal-→r0\vec{r}_{0} of a point of the plane and two linearly independent vectors u→normal-→u\vec{u} and v→normal-→v\vec{v}parallel to the plane: